好久没更新博客了，这段时间比较忙。以后争取有空多贴点好的问题，多谢大家光临我的博客。

Problem

If $a,b,c$ are positive real numbers such that $a+b+c=3$,then
$$\sqrt{a^3+3b}+\sqrt{b^3+3c}+\sqrt{c^3+3a}\geq 6$$

Proof. By the Cauchy-Schwarz inequality,we have
$$(a^3+3b)(a+3b)\geq (a^2+3b)^2$$
Thus,it suffices to show that
$$\sum\frac{a^2+3b}{\sqrt{a+3b}}\geq 6$$
By Holder inequality,we have
$$\left(\sum\frac{a^2+3b}{\sqrt{a+3b}}\right)^{2}\left[\sum(a^2+3b)(a+3b)\right]\geq \left[\sum(a^2+3b)\right]^3=\left(\sum{a^2}+9\right)^3$$
Therefore,it is enough to show that
$$\left(\sum a^2+9\right)^3\geq 36\sum(a^2+3b)(a+3b)$$
Let $p=a+b+c=3$ and $q=ab+bc+ca, q\leq 3$.We have
$$\sum a^2+9=p^2-2q+9=2(9-q)$$
$$\begin{align*} \sum(a^2+3b)(a+3b)&=\sum a^3+3\sum a^2b+9\sum a^2+3\sum ab\\ &=(p^3-3pq+3abc)+3\sum a^2b+9(p^2-2q)+3q \\ &=108-24q+3\left(abc+\sum a^2b \right)  \end{align*}$$
Since
$$a^2b+b^2c+c^2a+abc\leq \frac{4}{27}(a+b+c)^3$$
We get
$$\sum{(a^2+3b)(a+3b)}\leq 24(5-q)$$
Thus.it suffices to show that
$$(9-a)^3\geq 108(5-a)$$
Which is equivalent to the obvious inequality
$$(3-q)^2(21-q)\geq 0$$
The equality holds for $a=b=c=1$.